Question: $\dfrac{ 3q - 4r }{ -4 } = \dfrac{ -9q - s }{ 2 }$ Solve for $q$.
Solution: Multiply both sides by the left denominator. $\dfrac{ 3q - 4r }{ -{4} } = \dfrac{ -9q - s }{ 2 }$ $-{4} \cdot \dfrac{ 3q - 4r }{ -{4} } = -{4} \cdot \dfrac{ -9q - s }{ 2 }$ $3q - 4r = -{4} \cdot \dfrac { -9q - s }{ 2 }$ Reduce the right side. $3q - 4r = -{4} \cdot \dfrac{ -9q - s }{ {2} }$ $3q - 4r = -{2} \cdot \left( -9q - s \right)$ Distribute the right side $3q - 4r = -{2} \cdot \left( -{9q} - {s} \right)$ $3q - 4r = {18}q + {2}s$ Combine $q$ terms on the left. ${3q} - 4r = {18q} + 2s$ $-{15q} - 4r = 2s$ Move the $r$ term to the right. $-15q - {4r} = 2s$ $-15q = 2s + {4r}$ Isolate $q$ by dividing both sides by its coefficient. $-{15}q = 2s + 4r$ $q = \dfrac{ 2s + 4r }{ -{15} }$ Swap signs so the denominator isn't negative. $q = \dfrac{ -{2}s - {4}r }{ {15} }$